Question

If the position vectors of the vertices $A,B$ and $C$ of a $\Delta ABC$ are respectively $4\hat{i}+7\hat{j}+8\hat{k},2\hat{i}+3\hat{j}+4\hat{k}$ and $2\hat{i}+5\hat{j}+7\hat{k}$ , then the position vector of the point, where the bisector of $\angle A$ meets BC is :

• A. $\frac{1}{2}\left(4\hat{i}+8\hat{j}+11\hat{k}\right)$
• B. $\frac{1}{3}\left(6\hat{i}+11\hat{j}+15\hat{k}\right)$
• C. $\frac{1}{3}\left(6\hat{i}+13\hat{j}+18\hat{k}\right)$
• D. $\frac{1}{4}\left(8\hat{i}+14\hat{j}+19\hat{k}\right)$

Answer 1

Answer: (C)

From given data, we plot the graph below

$D$ is the point of intersection of internal angle biscctor of $A$ with line $BC$.
Now, $\overline{AB}=(2\hat{i}+3\hat{j}+4\hat{k})-(4\hat{i}+7\dot{j}+8\hat{k})=-2\hat{i}-4\hat{j}-4\hat{k}$
And $\overrightarrow{AC}=(2\hat{i}+5\hat{j}+7\hat{k})-(4\hat{i}+7\hat{j}+8\hat{k})=-2\hat{i}-2\hat{j}-\hat{k}$
$|A\vec{B}|=6,|\overrightarrow{AC}|=3$
$\frac{BD}{CD}=\frac{AB}{AC}=\frac{6}{3}=\frac{2}{1}$
Therefore, position vector of $D$ Point is
$\frac{(2(2i+5j+7k)+1(2i+3j+4k))}{3}=\frac{1}{3}(6\hat{i}+13\hat{j}+18\dot{k})$